Hence the maximum pressure exerted by the brick is 2450 dyne/cm²
Given the mass of the brick=5kg
Dimension of the brick=20cm*10cm*2cm.
So weight of the brick= F= mg
Where f= force on the body due to gravity
where m= mass, g= gravity.
So F= 5000*9.8
by doing the multiplication we get the result as
F= 49000 dyne where dyne is unit of force
Where surface area is 10cm*2cm is in contact with the ground then
Area= (10*2)= 20cm²
Applying pressure formula = F/Area= 49000/20
By dividing we get the answer as= 2450 dyne/cm²
When surface area is 20cm*2cm is in contact then
Area= 20cm*2cm= 40cm²
Applying pressure formula = F/Area= 49000/40 dyne/cm²= 1225 dyne/cm²
When the surface area is 20cm*10cm in contact then
Area= 20cm*10cm= 200cm²
By applying the pressure formula= F/Area= 49000/200=245 dyne/cm²
Hence the maximum pressure exerted by the brick is 2450 dyne/cm²
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A 75.2 kg base runner begins his slide into second base while moving at a speed of 4.17 m/s. He slides so that his speed is zero just as he reaches the base. The acceleration of gravity is 9.8 m/s². What is the magnitude of the mechanical energy lost due to friction acting on the run- ner? Answer in units of J.
The mechanical energy loss magnitude due to friction acting on the slider.
What is the mechanical energy loss due to rotor friction?To calculate mechanical energy use the following formula: Mechanical energy = ½ mv2 + mgh. h is the height from the floor. From this formula we can see that the only variables are mass, altitude and velocity. Power = power x speed. Friction force multiplied by constant velocity tells you how much energy is "lost" in one second. If you know the length of the lamp, you can combine that with a constant velocity to find out how many seconds it took to dissipate the amount of energy.Mechanical energy is the ratio of the speed at which the force is applied to the machine to the speed at which the load moves. Mechanical energy is also defined as the ratio of force displacement to load displacement.To learn more about Mechanical energy from the given link:
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an athlete suspected of using steroids is given two tests that operate independently of each other. test a has probability 0.9 of being positive if steroids have been used. test b has probability 0.8 of being positive if steroids have been used.
P(Event) is a probability that an event will occur, and conditional probability may be used to compute it. Given that an athlete has taken steroids, the likelihood that they will test positive for steroid usage is 0.72, or 72%.
If steroids have been used, Test A has a 0.9 chance of being positive, hence P(Test A Positive | Steroids Used) = 0.9
If steroids have been used, there is a 0.8 likelihood that Test B will be positive, hence P(Test B Positive | Steroids Used) = 0.8
The likelihood that a sportsperson will use steroids is denoted by P. (Steroids Used).
The probability that an athlete will test positive for steroid use can be determined using the conditional probability formula, which is as follows: P(Test Positive | Steroids Used) = P(Test A Positive and Test B Positive | Steroids Used) = P(Test A Positive | Steroids Used) * P(Test B Positive | Steroids Used).
= 0.9 * 0.8 = 0.72
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which of the following is the best way driver/operators can prevent the aerial device from moving or chattering along a frozen solid surface on an angle? group of answer choices using wheel chocks deploying manual stabilizers positioning the apparatus laterally positioning the apparatus longitudinally
Using wheel clocks is the best option.
To stop an aerial device from moving or chattering on an angled frozen solid surface, wheel chocks are intended to be installed in front of and behind the wheels of the aerial device. The aerial equipment can't move or chatter since the chocks mechanically stop the wheels from rotating. Additionally, they can be utilised to keep the apparatus stable on the surface by adding an additional layer of friction between the ground and the object. Wheel chocks are an efficient way to stop aerial equipment from shifting or chattering on an angled frozen solid surface when used in conjunction with hand stabilisers.
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Which of the following will have the highest surface tension?
A
Water
B
Ethanol
C
Methanol
D
Ethanal
Water has the highest surface tension than ethanol, methanol, or ethanal.
The cohesive nature of a liquid's molecules gives rise to a property of the liquid's surface known as surface tension, which enables the liquid to resist the action of an outside force. After mercury, water has the highest surface tension of any liquid. This is because hydrogen bonds are present in water molecules, which gives water its high surface tension. As a result of water's surface tension, water molecules that are in close proximity to one another at the liquid's surface (where they are in contact with air) stick together to form an invisible film. When water is heated to 25 degrees Celsius, its surface tension is 72 dynes/cm. To crack a film of water on the surface that is one centimeter long and requires 72 dynes of force.
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a particle of uniform mass m is locaed inside a uniform solid sphere of radius r and mass m at a distance r from its center (a) sho that the gravitational potential energy of the system is U=GMmr^2 /2R^3 - 3GmM/2R ?
(a) The gravitational potential energy of the system is [tex]\frac{GmM}{2R^3}r^2 - \frac{3GmM}{2R}[/tex] (Proved), and (b) the amount of work done by the gravitational force in bringing the particle from the surface of the sphere to its center will [tex]-\frac{GMm}{2R}[/tex].
Assuming that a concentric spherical surface passes through where a particle of mass m resides.
so from that imaginary boundary
The mass of the inner part is
M' = [tex]\frac{M}{\frac{4\pi r^3}{3} } . \frac{4}{3} \pi r^3[/tex]= Mr³/r³
The potential at P due to this inner part is
V₁ = -GM'/r
or V₁ = -GMr²/R³
Now, to find the potential at P due to the outer part of the sphere, we will divide this by concentric shells.
so, dm = [tex]\frac{M}{\frac{4\pi r^3}{3} } . \frac{4}{3} \pi x^2 dx[/tex] = 3Mx²dx/R³
by further solving,
[tex]-\frac{Gdm}{x} = -3\frac{GM}{R^3}xdx[/tex]
so by integrating,
V₂ = [tex]\int\limits^R_r -3\frac{GM}{R^3}xdx[/tex]
or V₂ = [tex]-\frac{3GM}{2R^3} (R^2-r^2)[/tex]
So, the total potential at P will be
V = V₁ + V₂ = -GMr²/R³ [tex]-\frac{3GM}{2R^3} (R^2-r^2)[/tex]
or V = [tex]-\frac{GM}{2R^3}(3R^2-r^2)[/tex]
and we know that potential energy U = mV = [tex]\frac{GmM}{2R^3}r^2 - \frac{3GmM}{2R}[/tex] {Hence Proved}
(b) we know that potential energy at the center is Uc = [tex]-\frac{3GmM}{2R}[/tex] {r=0}
At surface {r+R} Us = [tex]-\frac{GMm}{R}[/tex]
hence, work done W = Uc - Us = [tex]-\frac{GMm}{2R}[/tex]
Therefore, (a) The gravitational potential energy of the system is [tex]\frac{GmM}{2R^3}r^2 - \frac{3GmM}{2R}[/tex] (Proved), and (b) the amount of work done by the gravitational force in bringing the particle from the surface of the sphere to its center will be [tex]-\frac{GMm}{2R}[/tex].
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The complete question is:
a particle of uniform mass m is located inside a uniform solid sphere of radius r and mass m at a distance r from its center (a) sho that the gravitational potential energy of the system is U=GMmr^2 /2R^3 - 3GmM/2R ? (b) Write an expression for the amount of work done by the gravitational force in bringing the particle from the surface of the sphere to its center.
The resultant of a 40-N force at right angles to a 30-N force is
Answer:
The magnitude of the resultant force of the two forces (40 N and 30 N) at right angles is 50 N.
Explanation:
The resultant of two forces at right angles to each other is found by using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the longest side, opposite the right angle) is equal to the sum of the squares of the other two sides (the two shorter sides).
In this case, the two forces form the two shorter sides of the triangle and the resultant force is the hypotenuse.
So the magnitude of the resultant force can be calculated as:
√(40^2 + 30^2) = √(1600 + 900) = √2500 = 50 N
The magnitude of the resultant force of the two forces (40 N and 30 N) at right angles is 50 N.
during and after eyewall replacement, which of the following is true about the hurricane's strength?
Since a tropical cyclone's eyewall has the greatest winds, the storm often weakens during this time when the inner wall is "choked" by the outer wall.
A ring of tall thunderstorms that form the eyewall produce intense rains and typically have the greatest winds. Wind speed, a measure of the storm's strength, can alter due to changes in the eye and eyewall's structure. The cycle of eyewall replacement is one characteristic of major hurricanes. To put it simply, a new eye starts to form around the old one. The old eye is replaced with the new one, whose diameter gradually shrinks.
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A proton is fired horizontally into a 1.0×10^5 N/C vertical electric field. It rises 1.0 cm vertically after having traveled 7.0 cm horizontally. What was the proton's initial speed?
The initial horizontal speed of the proton is 0.16 m/s.
What is the time of motion of the proton?
The time of motion of the proton is calculated by applying the following kinematic equation.
t = √ ( 2gh )
where;
h is the height risen by the protong is the acceleration due to gravityt = √ ( 2 x 9.8 x 0.01 )
t = 0.44 second
The initial horizontal speed of the proton is calculated as;
d = Vₓt
Vₓ = d / t
Vₓ = ( 0.07 m ) / ( 0.44 s )
Vₓ = 0.16 m/s
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How far (in m) does a car going 21 m/s travel in 7.1 s?
Use the impulse-momentum theorem to find how long a stone falling straight down takes to increase its speed from 5.5 m/s to 10.4 m/s.
The stone will descend straight down for 0.5 seconds, increasing its speed from 5.5 m/s to 10.4 m/s.
Newton's second law of motion states that F = ma where ma is the object's mass and an is its acceleration.
You can also write the equation as:
F = m (v - u)/t
Ft =m (v-u)
mat = m (v-u)
at = v-u
t = v-u/a
Given the following parameters;
v = 10.4 m/s
u = 5.5 m/s
a = 9.8m/s²
Substitute the given parameters into the formula to have:
t = 10.4-5.5/9.8
t = 4.9/9.8
t = 0.5 sec
The stone's speed will therefore increase from 5.5 m/s to 10.4 m/s in 0.5 seconds as it drops directly to the ground.
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consider the following observations. classify each observation based on whether it is a real observation (a true statement of something we can actually see from earth) or one that is not real (a statement of something that does not really occur as seen from earth). view available hint(s)for part a
Real: True statements
Mercury undergoes a complete cycle of phases.Daily moon rises in the east and sets in the west.Stars circle daily around north or south celestial poleEach year, neighbouring stars' positions gradually oscillate back and forth.A distance galaxy rises in east, sets in west each dayNot real: False statements:
We sometimes see a crescent JupiterBeyond Saturn, a planet rises in the west and sets in the eastEarth observation is the gathering of information on the physical, chemical, and biological processes occurring on planet Earth using remote sensing technology, often via satellites carrying imaging equipment. Earth observation is used to monitor and assess environmental changes, both natural and man-made, as well as their state.
Seismic and Global Positioning System (GPS) stations, as well as floatable buoys for monitoring ocean currents, temperature, and salinity, are some of the techniques used today for Earth observation. Air quality and rainfall patterns are also recorded by land-based stations. Space-based technologies provide repeatable datasets that provide a unique opportunity to gather data about the world when combined with appropriate technique development and analysis.
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Following question may be like this:
Consider the following observations. Classify each observation based on whether it is a real observation (a true statement of something we can actually see from Earth) or one that is not real (a statement of something that does not really occur as seen from Earth).
- Mercury goes through a full cycle of phases
- Moon rises in east, sets in west each day
- stars circle daily around north or south celestial pole
- positions of nearby stars shift slightly back and forth each year
- a distance galaxy rises in east, sets in west each day
- we sometimes see a crescent Jupiter
- a planet beyond Saturn rises in west, sets in east
Forces at Work
The spring-loaded cart is a simple system, which can make it easier to
analyze and apply the scientific ideas observed to a larger, more
complex system. A skateboarder is another simple system. A
skateboarder would not get anywhere without forces. Since the
skateboard does not have a motor, the skateboarder must be the one
that supplies the power to make it go. But how does this push cause
the motion? Why does the skateboarder have to continually push the
skateboard to keep it in motion?
Similar cause-and-effect relationships involved in the spring-loaded
cart can be applied to your model of the fire extinguisher go-kart.
What cause-and-effect relationships identified in
the spring-loaded cart can be applied to the
rocket sled?
Effect
Cause
Click to
add text
Cause
Click to
add text
Click to
add text
Effect
Click to
add text
The skateboard was initially at rest, and by the time of Newton's first law, it would still be in that state. This law asserts that unless acted upon by an external or net force, a body will remain at rest or continue in a condition of uniform motion with constant velocity (in a situation where not only one force is in action).
According to this law, in order to move the skateboard, it needs an external source of force, which the skateboarder indirectly provides.
The skateboarder reverses forward on the pavement (that is he applies a force on the road in a direction opposite the direction of intended motion).
According to Newton's third law, the skateboarder's action results in an equal and oppositely directed reaction from the road on the skateboarder.
According to Newton's third law, action and response are equal but directed in opposite directions. As a result, the road pulls the skateboarder forward in response to this rearward force, which moves both the skateboarder and the skateboard in the same direction.
The acceleration of the cart is inversely related to its mass. As a result, the acceleration of the cart will decrease as its mass increases.
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common transparent tape becomes charged when pulled from a dispenser. if one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 13.0 mg piece of tape held 1.30 cm above another. (the magnitude of this charge is consistent with what is typical of static electricity.)
If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight.
calculate the magnitude of the charge if electrostatic force?
The magnitude of charges on the electron and proton are `1.6xx10^(-19)` C.M Mass of the electrons is `m_(e)=9.1xx10^(-31)` kg and mass of proton is `m_(p)=1.6xx10^(-27)` kg .n the equation Felect = k • Q1 • Q2 / d2 , the symbol Felect represents the electrostatic force of attraction or repulsion between objects 1 and 2.The force is perpendicular to both the velocity v of the charge q and the magnetic field B. 2. The magnitude of the force is F = qvB sinθ where θ is the angle < 180 degrees between the velocity and the magnetic field.This force emerges from the interaction between two charged objects (or point charges) and its magnitude is calculated by F=kQ1Q2r2 F = k Q 1 Q 2 r.
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What are 2 contradictions of inertia
A body's inertia is a passive characteristic that only allows it to oppose active agents like forces and torques. A moving body continues to move not because of its inertia but rather because there is no external force present to cause it to slow down, veer off course, or accelerate.
What is inertia?The idea of inertia states that an object will keep moving in the same direction until another force causes it to change its direction or speed.
It is important to recognise that Newton used the slang phrase "the principle of inertia" when he discussed it in his first law of motion.
resistance of the body to changes in momentum
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A net force is required to stop a 100 kg car , traveling at 3 m.s -1 . By what factor would the NT force acting on the car change if
a) the car was stopped in double time
b) the car was stopped in 1 tenth of the time
c) the car was traveling at double the velocity and it's stopped in the same time ?
You must add up all of the forces operating on a certain object in order to calculate the net force acting on that object.
How are net forces calculated?These interactions can be summed up by Newton's second law of motion. Net Force = Mass Acceleration, or F = m a, is how the preceding equation for acceleration may be expressed to solve for net force.For the 1000kg automobile to accelerate by 3m/s2, 3000N of force is required.Numerous additional forces, such as gravitational force Fg, frictional force Ff, and the normal force, act on the body when a force is applied to it. These additional forces serve to balance the original force. FNet = Fa + Fg + Ff + FN is the result, and it represents the net force formula.To learn more about net force refer to:
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two identical spheres with mass are hung from silk threads of length (fig. p21.60). the spheres have the same charge, so the radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. show that if the angle is small, the equilibrium separation between the spheres is (hint: if is small, then )
The equilibrium distance between the centers of the balls will be
x =[tex](2q1q2lmg)^{\frac{1}{3} }[/tex].
What are the main forces acting on each balll?To find the equilibrium distance between the two balls, we need to consider the forces acting on each ball. The main forces acting on each ball are the gravitational force and the electrostatic force between the two balls.
The gravitational force on each ball is given by:
Fg = mg (where g is the acceleration due to gravity)
The electrostatic force between the two balls is given by:
Fe = kq1q2/x² (where k is Coulomb's constant, and x is the distance between the centers of the two balls)
For equilibrium to occur, the net force on each ball must be zero. Therefore, we can set up the following equation:
Fg - Fe = 0
Substituting in the values for Fg and Fe, we get:
mg - (kq1q2/x²) = 0
Solving for x, we get:
x =[tex](2q1q2lmg)^{\frac{1}{3} }[/tex].
where l is the length of the silk threads.
So, the equilibrium distance between the centers of the balls will be:
x =[tex](2q1q2lmg)^{\frac{1}{3} }[/tex].
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An airfoil has upper and lower surfaces defined by the equations: 1. For 0 < (x/c) < 0.6 Y upper = 0.4(x/c), Y lower = 0.2(x/c) 2. For 0.6 < (x/c) < 1.0
Y upper = 0.6(1 - x/c), Y lower = 0.3(1 - x/c) Note you need to do the analysis of over each segment separately. Analyze this using the thin airfoil theory, 1. Show that the equations for the camber and half thickness distributions are A. For 0 < (x/c) < 0.6 n = 0.3(x/c), n = 0.1(x/c) B. For 0.6 < (x/c) < 1.0
n = 0.45 (1 - x/c), n = 0.15 (1 - x/c)
The equations for the camber and half thickness distributions are:
A. For 0 < (x/c) < 0.6 n = 0.3(x/c), t = 0.1(x/c)
B. For 0.6 < (x/c) < 1.0 n = 0.45 (1 - x/c), t = 0.15 (1 - x/c)
The camber and half thickness distributions can be found using the thin airfoil theory. According to this theory, the camber and half thickness distributions are given by:
n = (Y upper - Y lower ) / 2 (for camber)
t = (Y upper + Y lower ) / 2 (for half thickness)
For the given airfoil, we can find the camber and half thickness distributions by applying the above equations to each segment separately.
The theory simply states that flow around an airfoil is described as two-dimensional flow around a thin airfoil. It is possible to imagine it addressing an airfoil with zero thickness and infinite wingspan.
For 0 < (x/c) < 0.6
Y upper = 0.4(x/c), Y lower = 0.2(x/c)
Therefore,
n = (0.4(x/c) - 0.2(x/c)) / 2 = 0.3(x/c)
t = (0.4(x/c) + 0.2(x/c)) / 2 = 0.1(x/c)
For 0.6 < (x/c) < 1.0
Y upper = 0.6(1 - x/c), Y lower = 0.3(1 - x/c)
Therefore,
n = (0.6(1 - x/c) - 0.3(1 - x/c)) / 2 = 0.45 (1 - x/c)
t = (0.6(1 - x/c) + 0.3(1 - x/c)) / 2 = 0.15 (1 - x/c)
So, the equations for the camber and half thickness distributions are:
A. For 0 < (x/c) < 0.6 n = 0.3(x/c), t = 0.1(x/c)
B. For 0.6 < (x/c) < 1.0 n = 0.45 (1 - x/c), t = 0.15 (1 - x/c)
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A stone is thrown downwards with a speed of 5ms, from the top of a building of height 180 m. Take point of projection as the origin and vertical downward direction as negative. Determine
a. the velocity of stone at the end of 3 s.
b. the displacement of stone at the end of 3 s.
c. the time taken by the stone to travel a distance of 150 m.
Explanation:
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the velocity of a 1.7 kg block sliding down a frictionless inclined plane is found to be 1.54 m/s. 1.70 s later, it has a velocity of 6.13 m/s.
The angle of plane is 15 degrees.
mg x Sin A gives the force that pushes the mass down the incline.
Additionally, this force uses the same formula m x a which is mass times acceleration,
where a is the original acceleration in the plane.
The block sliding has a mass of 1.7 kg and an initial speed of 1.54 m/s.
the velocity is 6.13 m/s after the time of 1.70 seconds, which is given.
By dividing the difference between the two initial velocities by the given time, or
(6.13 m/s - 1.54 m/s) / 1.70s = 2.70 [tex]m/s^2[/tex],
the acceleration a may be calculated.
ma is therefore which is equal to mass time acceleration g times sin A, or
sin A = acceleration (a) / gravity (g) , or
2.70 [tex]m/s^2[/tex] times 9.81 times 0.275.
A is equal to arc sin (0.275) = or sin A = 15 degrees
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Force (N)
896760 50 3NTO
10
2
I
0 0.04 0.08 0.12 0.16 0.20
Extension (m)
13. Using the graph above, (a) determine the spring constant, (b) write the Hooke's law equation for this
spring.
a) The spring constant is 896760 N/m.
b) F = -896760x
How do we arrive at the values?(a) To determine the spring constant, we need to find the slope of the linear portion of the graph. The slope can be found by dividing the change in force by the change in extension.
(b) Hooke's law states that the force applied to a spring is proportional to the spring's extension, with the constant of proportionality being the spring constant. So the equation for this spring would be:
F = kx
where F is the force, x is the extension, and k is the spring constant.
Therefore, the spring constant is 896760 N/m. It could then be concluded that the spring constant is as given above.
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In which of the following situations is the velocity constant, but the acceleration is not zero? In no situation can velocity be constant at the same time there is non-zero acceleration A car drives at constant speed along a winding road. A sprinter runs a 100-meter dash along a straight track.A train moves along a straight track at constant speed
In no case can velocity be constant while there is non-zero acceleration.
What is the relation between velocity and acceleration?During uniformly accelerated, straight-line motion, the relationship between velocity and time is straightforward. The greater the change in velocity, the longer the acceleration. When the acceleration is constant, the change in velocity is directly proportional to time.This is also the case in uniform motion, where the object moves at a constant speed. Because acceleration implies a non-zero change in velocity with respect to time, a particle can have zero acceleration if the velocity is constant but non-zero.When the magnitude and direction of a velocity do not change over time, it is said to be constant. In other words, this is when the rate of change of an object's position remains constant over time.To learn more about acceleration refer to :
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The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s= 2 sin π t + 3 cos π t, where t is measured in seconds.
a. Find the average velocity during each time period:
i (1, 2)
ii. (1, 1.1)
iii. (1, 1.01)
iv. (1, 1.001)
b. Estimate the instantaneous velocity of the particle when t=1.
(a.i) The average velocity of the particle for period (1, 2) is 0 cm/s.
(a.ii) The average velocity of the particle for period (1, 1.1) is - 3.1 cm/s.
(a.iii) The average velocity of the particle for period (1, 1.01) is -3.3 cm/s.
(a.iv) The average velocity of the particle for period (1, 1.001) is -3 cm/s.
(b) The instantaneous velocity of the particle when t=1 is - 2 cm/s.
What is average velocity?The average velocity of an object is the ratio of the total displacement to total time of motion of the object.
v = ( total displacement ) / ( total time )
v = ( s ) / ( t )
The average velocity of the particle for period (1, 2)
s = 2 sin π (1) + 3 cos π (1) + 2 sin π (2) + 3 cos π (2)
s = ( 0 - 3 ) + ( 0 + 3 ) = 0 cm
v = ( 0 cm ) / (1 + 2 ) = 0 cm/s
The average velocity of the particle for period (1, 1.1)
s = 2 sin π (1) + 3 cos π (1) + 2 sin π (1.1) + 3 cos π (1.1)
s = ( 0 - 3 ) + ( -0.62 - 2.85 ) = -6.47 cm
v = ( -6.47 cm ) / ( 1 + 1.1 )
v = -3.1 cm/s
The average velocity of the particle for period (1, 1.01)
s = 2 sin π (1) + 3 cos π (1) + 2 sin π (1.01) + 3 cos π (1.01)
s = ( 0 - 3 ) + ( -0.63 - 3 ) = -6.63 cm
v = ( -6.63 cm ) / ( 1 + 1.01 )
v = -3.3 cm/s
The average velocity of the particle for period (1, 1.001)
s = 2 sin π (1) + 3 cos π (1) + 2 sin π (1.001) + 3 cos π (1.001)
s = ( 0 - 3 ) + ( -0.0063 - 3 ) = -6.0063 cm
v = ( -6.0063 cm ) / ( 1 + 1.001 )
v = -3 cm/s
The instantaneous velocity of the particle when the time = 1, is calculated as follows;
v (t) = ds / dt
v (t) = 2 cosπ t - 3 sin π t
v (1) = 2 cosπ (1) - 3 sin π (1)
v ( 1 ) = -2 cm /s
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Why is B the correct answer? Thanks :)
The final speed is v and the change in the total kinetic energy is 6mv² (Option B)
How do I determine the final velocity?The final speed after the collision can be obtained as illustrated below:
Mass of 1st object (m₁) = 2mspeed of 1st object (u₁) = 3vMass of 2nd object (m₂) = 4mspeed of 2nd object (u₂) = 0Final speed (v₀) = ?Momentum before = momentum after
m₁u₁ + m₂u₂ = v₀(m₁ + m₂)
Divide both sides by (m₁ + m₂)
v₀ = [m₁u₁ + m₂u₂] / (m₁ + m₂)
v₀ = [(2m × 3v) + (4m × 0)] / (2m + 4m)
v₀ = [6mv + 0] / 6m
v₀ = 6mv / 6m
v₀ = v
Thus, the final speed is v
How do I determine the change in the kinetic energy?First, we shall determine the total initial kinetic energy. Details below:
Mass of 1st object (m₁) = 2mspeed of 1st object (u₁) = 3vMass of 2nd object (m₂) = 4mspeed of 2nd object (u₂) = 0Total Initial Kinetic energy (KE₁) =?KE₁ = ½m₁u₁² + ½m₂u₂²
KE₁ = [½ × 2m × (3v)²] + [½ × 2m × 0²]
KE₁ = [m × 9v²] + 0
KE₁ = 9mv²
Next, we shall determine the total final kinetic energy. Details below:
Total mass (m) = 2m + 4m = 6mFinal speed (v₀) = vTotal final Kinetic energy (KE₂) =?KE₂ = ½mv²
KE₂ = ½ × 6m × v²
KE₂ = 3mv²
Finally, we shall determine the change in the kinetic energy of the car can be obtained as follow:
Total Initial Kinetic energy (KE₁) = 9mv²
Total final Kinetic energy (KE₂) = 3mv²
Change in total kinetic energy (ΔKE) =?
ΔKE =KE₁ - KE₂
ΔKE = 9mv² - 3mv²
ΔKE = 6mv²
In conclusion,
Final speed = vChange in total kinetic energy = 6mv²Thus, the correct answer is Option B
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Compare accuracy and precision.
Answer:
Accuracy refers to how close a measurement is to the true or accepted value. Precision refers to how close measurements of the same item are to each other. Precision is independent of accuracy.
Answer:
Accuracy refers to how close a measurement is to the true or accepted value. Precision refers to how close measurements of the same item are to each other.
Explanation:
Precision is independent of accuracy.
A rabbit escaping a fox runs across level ground, zipping north for 3.04 m, darting exactly northwest for 2.56 m, and then dropping 0.6 m straight down underground into the safety of its burrow.How far, d, does the rabbit end up from its starting point?
Zipping 3.04 metres distance in north, precisely 2.56 metres northwest, and finally plunging 0.6 metres straight down into the security of its burrow 4.59 m later, the rabbit has travelled from its starting place.
The rabbit initially travels 3.04 metres north, then 2.56 metres northwest. The Pythagorean theorem may be used to determine the hypotenuse of a right triangle with 3.04 m and 2.56 m-long legs in order to determine the total distance the rabbit has travelled in a north-westerly direction:
d = sqrt(3.04^2 + 2.56^2)
sqrt(15.984), sqrt(9.4304 + 6.5536), and d = 3.99 m
The vertical component of the rabbit's distance travelled must now be included. The rabbit descended 0.6 metres below earth. The rabbit travels a total distance of d = sqrt(15.984) + 0.6 m d = 3.99 m + 0.6 m = 4.59 m from its starting place.
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If the number of turns on the secondary coil of a transformer are less than those on the primary, the result is a.
A) step-up transoformer.
B) a dc transformer.
C) 120-v transformer
D) step-down transoformer
E) 220-v transformer
A certain amount of heat is added to some water so that its temperature rises. The same amount of heat is added to a piece of alu- minum with the same mass as the water. Which has the higher temperature change? A)water B) aluminum C) they gave equal temperature changes
At upper temperatures, the thermal conductivity decreases slowly with increasing temperature than depicted by this equation. Aluminum has the higher temperature change.
What is Aluminum ?Aluminum is a silvery-white metal and is the 13th element on the periodic table. A surprising fact about aluminum is that it is the most common metal on earth.Aluminum (Al), also known as aluminum, chemical element, light silvery-white metal of the 13th main group (III a, or boron group) of the periodic table. Aluminum is the most abundant metallic element in the earth's crust and the most commonly used non-ferrous metal. Aluminum is a silver-white colored lightest metal. Soft and easy to play. Aluminum is used in a variety of products including cans, foils, kitchenware, window frames, beer kegs and aircraft components.to learn more about Aluminum from the given link:
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A boy throws a ball straight up into the air. It reaches the highest point of its flight after 4 seconds. How fast was the ball going when it left the boy's hand?
The ball's exit velocity from the boy's hand was 39.4 m/s.
The time it took the ball to reach its highest point is 4 seconds, which is the explanation.
Use the first equation of motion, v = u + gt, to determine the ball's initial velocity when it left the boy's hand.
Initial velocity (v) is used here. terminal velocity is u. t is the amount of time, and g is the gravitational acceleration, with a value of 9.8 m/s2.
At its highest point, the ball's final velocity, v, equals zero.
When the supplied values are substituted in the equation above, we get 0=u-9.8m/s2.
Therefore, the ball's exit velocity from the boy's hand was 39.4 m/s.
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as infants extract an increasing number of potential word forms from the speech stream they hear, they begin to associate these with concrete, perceptually available objects in their world. this is referred to as multiple choice question.
Infants extracting an increasing number of potential word forms from the speech stream they hear, begin to associate these with concrete, perceptually available objects is referred to as statistical learning.
Statistical learning refers to the process by which infants extract potential word forms from the speech stream they hear and begin to associate them with concrete, perceptually available objects. This process is thought to be a crucial component of language development, as it allows infants to begin to understand the structure and meaning of the language they are exposed to.
Through statistical learning, infants are able to identify patterns and regularities in the speech they hear, such as the relationship between a word and the object it refers to. This ability enables them to begin to build a vocabulary and develop an understanding of the grammar and syntax of their native language. Additionally, statistical learning may also play a role in the development of other cognitive abilities, such as memory and attention.
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Your question seems incomplete, but I suppose the question was:
"As infants extract an increasing number of potential word forms from the speech stream they hear, they begin to associate these with concrete, perceptually available objects in their world. this is referred to as"
Calculate the first and second velocities of the car with one washer attached to the pulley, using the formulas
v1 = 0.25 m / t1, and
v2 = 0.25 m / (t2 – t1)
where t1 and t2 are the average times the car took to reach the 0.25 and the 0.50 meter marks. Record these velocities, rounded to two decimal places, in Table E.
What is the first velocity of the car with one washer at the 0.25 meter mark?
m/s
What is the second velocity of the car with one washer at the 0.50 meter mark?
m/s
The speed is unusual (65 [tex]\frac{mi}{h}[/tex] + 55[tex]\frac{mi}{h}[/tex] ), The two rates were not maintained keep up with the times.
Calculating the problem:
The speed changed once a certain amount of time had passed. Speed = We get 65 = [tex]\frac{mi}{h}[/tex] = [tex]\frac{130 mi}{h}[/tex] by substituting variables, which is the formula for this = [tex]\frac{d}{t}[/tex]
In the interim, t1 has a value of 2.0 hours.
At lower speeds, the time is shown as t2 = T - t1 = 3.33 h — 2.0 h = 1.33 h.
The difference between the two speeds, such as speed = 73 miles, is also shown in this.
D is the sum of d1 and d2, which equals 203 miles.
The normal speed is = [tex]\frac{d}{t}[/tex] = [tex]\frac{203mi}{h}[/tex]= 61.
The two rates were not kept up with the times because of the unusual speed (65[tex]\frac{mi}{h}[/tex] + 55 [tex]\frac{mi}{h}[/tex]).
What is speed?
The direction in which a body or object is moving is determined by velocity.
Most of the time, velocity is a scalar number. In essence, velocity is a vector quantity. The term "velocity" refers to the rate of change in displacement and distance. speed is indicated by the speed and direction of travel (for instance, 60 kilometers per hour north).
Typically, velocity is defined as the vector value of the speed and motion direction. The speed at which something moves in one direction is known as velocity. Velocity can be used to measure both the speed of a car traveling north on a highway and the speed of a rocket taking off into space.
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Answer:
What is the first velocity of the car with one washer at the 0.25 meter mark?
1st = 0.11 m/s
2nd = 0.13 m/s
3rd = 0.19 m/s
What is the second velocity of the car with one washer at the 0.50 meter mark?
1st = 0.28 m/s
2nd = 0.36 m/s
3rd = 0.45 m/s
Explanation:
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